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1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
\(\left(1+\dfrac{x+1}{17}\right)+\left(1+\dfrac{x+2}{16}\right)=\left(1+\dfrac{x+3}{15}\right)+\left(1+\dfrac{x+4}{14}\right)\)
\(\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì : \(\dfrac{1}{17}< \dfrac{1}{15};\dfrac{1}{16}< \dfrac{1}{14}\Rightarrow\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}< 0\)
\(\Rightarrow x+18=0\Rightarrow x=0-18=-18\)
Từ đầu đến đến dòng thứ tư thì mình đồng ý, nhưng mình nghĩ không nhất thiết phải so sánh, chỉ cần làm tiếp như sau:
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+18=0\\\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}=0\end{matrix}\right.\)
rồi tính tiếp là OK rồi. Dù gì thì cũng xin cảm ơn nha ;)
\(A=1\dfrac{1}{15}.1\dfrac{1}{16}.1\dfrac{1}{17}......1\dfrac{1}{2016}.1\dfrac{1}{2017}\)
\(A=\dfrac{16}{15}.\dfrac{17}{16}.\dfrac{18}{17}......\dfrac{2017}{2016}.\dfrac{2018}{2017}\)
\(A=\dfrac{16.17.18......2017.2018}{15.16.17......2016.2017}\)
\(A=\dfrac{2018}{15}\)
Bài 1:
a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)
\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)
c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)
\(\dfrac{-8}{13}+\dfrac{-7}{17}+\dfrac{21}{13}\le x\le\dfrac{-9}{14}+3+\dfrac{5}{-14}\)
=> \(\dfrac{10}{17}\le x\le2\)
=> \(\dfrac{10}{17}\le\dfrac{17x}{17}\le\dfrac{34}{17}\)
=> 10 \(\le17x\le34\)
=> x = 1; 2 (thỏa mãn)
@Khánh Linh
Ta có: \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Rightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Rightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Rightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Rightarrow\left(x+18\right).\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\) (1)
Mà \(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\ne0\) (2)
Từ (1) và (2) => x+18=0 => x=-18
Vậy x=-18
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
\(5\dfrac{8}{17}:x+\left(\dfrac{-4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
\(\Leftrightarrow\left(5\dfrac{8}{17}+\dfrac{-4}{17}\right):x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)
\(\Leftrightarrow\left(\dfrac{43}{17}+\dfrac{-4}{17}\right):x+\dfrac{22}{7}.\dfrac{3}{52}=\dfrac{4}{11}\)
\(\Leftrightarrow\dfrac{39}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)
\(\Leftrightarrow\dfrac{39}{17}:x=\dfrac{4}{11}-\dfrac{33}{182}\)
(cứ thế mà làm tiếp nhé, mình thấy số xấu quá)
9/17.x+268/17.x-345/17.x=16
(9/17+268/17-345/17).x=16
-4.x=16
x=16:-4
x=-4
\(\dfrac{9}{17}x+15\dfrac{13}{17}x-20\dfrac{5}{17}x=16\)
\(\left(\dfrac{9}{17}+15\dfrac{13}{17}-20\dfrac{5}{17}\right)x=16\)
\(\left(\dfrac{9}{17}+\dfrac{268}{17}-\dfrac{345}{17}\right)x=16\)
\(x=16:\left(\dfrac{9}{17}+\dfrac{268}{17}-\dfrac{345}{17}\right)\)
\(x=-4\)